§8.2 Quadratic Equation Apps Chabot Mathematics §8.2 Quadratic Equation Apps Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

8.2 Review § Any QUESTIONS About Any QUESTIONS About HomeWork MTH 55 Review § Any QUESTIONS About §8.2 → Quadratic Formula Any QUESTIONS About HomeWork §8.2 → HW-39

§8.2 Quadratic Formula The Quadratic Formula Problem Solving with the Quadratic Formula

This is one of the MOST FAMOUS Formulas in all of Mathematics The Quadratic Formula The solutions of ax2 + bx + c = 0 are given by This is one of the MOST FAMOUS Formulas in all of Mathematics

Example Circular WalkWay A Circular Pond 10 feet in diameter is to be surrounded by a “Paver” walkway that will be 2 feet wide. Find the AREA of the WalkWay Familiarize: Recall the Formula for the area, A ,of a Circle based on it’s radius, r Also the diameter, d, is half of r. Thus A in terms of d

Example Circular WalkWay Familiarize: Make a DIAGRAM Translate: Use Diagram of Subtractive Geometry = −

Example Circular WalkWay Translate: Diagram to Equation = −

Example Circular WalkWay CarryOUT: Solve Eqn for Awalk Using π ≈ 3.14 find

Example Circular WalkWay Check: Use Acircle = πr2  State: The Area of the Paver Walkway is about 75.4 ft2 Note that UNITS must be included in the Answer Statement

Example  Bike Tire BlowOut Devon set out on a 16 mile Bike Ride. Unfortunately after 10 miles of Biking BOTH tires Blew Out. Devon Had to complete the trip on Foot. Devon biked four miles per hour (4 mph) faster than than she walked The Entire journey took 2hrs and 40min Find Devon’s Walking Speed (or Rate)

Example  Bike Tire BlowOut Familiarize: Make diagram LET w ≡ Devon’s Walking Speed Recall the RATE Equation for Speed Distance = (Speed)·(Time)

Example  Bike Tire BlowOut Translate: The Biking Speed, b, is 4 mph faster than the Walking Speed → From the Diagram note Distances by Rate Equation: Biking Distance = 10 miles = b·tbike Walking Distance = 6 miles = w·twalk

Example  Bike Tire BlowOut Translate: Now the Total Distance of 16mi is the sum of the Biking & Walking Distances → From the Spd Eqn: Time = Dist/Spd

Example  Bike Tire BlowOut Translate: Thus by Speed Eqn: Next, the sum of the Biking and Walking times is 2hrs & 40min = 2-2/3hr →

Example  Bike Tire BlowOut CarryOut: Clear Fractions in the last Eqn by multiplying by the LCD of 3·(w+4) ·w:

Example  Bike Tire BlowOut CarryOut: Divide the last Eqn by 8 to yield the a Quadratic Equation This Quadratic eqn does NOT factor so use Quadratic Formula with a = 1, b = −2, and c = −9

Example  Bike Tire BlowOut CarryOut: find w by Quadratic Formula So Devon’s Walking Speed is

Example  Bike Tire BlowOut CarryOut: Since SPEED can NOT be Negative find: Check: Test to see that the time adds up to 2.67 hrs 

Example  Bike Tire BlowOut State: Devon’s Walking Speed was about 4.16 mph (quite a brisk pace)

Example  Partition Bldg A rectangular building whose depth (from the front of the building) is three times its frontage is divided into two parts by a partition that is 45 feet from, and parallel to, the front wall. If the rear portion of the building contains 2100 square feet, find the dimensions of the building. Familiarize: REALLY needs a Diagram

Example  Partition Bldg Familiarize: by Diagram Now LET x ≡ frontage of building, in feet. Translate: The other statements into Equations involving x

Example  Partition Bldg The Bldg depth is three times its frontage, x → 3x = depth of building, in feet The Bldg Depth is divided into two parts by a partition that is 45 feet from, and parallel to, the front wall → 3x – 45 = depth of rear portion, in ft

Example  Partition Bldg Now use the 2100 ft2 Area Constraint Area of rear = 2100 Area = x(3x−45), so

Example  Partition Bldg So x is either 35ft or −20ft But again Distances can NOT be negative Thus x = 35 ft Check: Use 2100 ft2 Area 

Example  Partition Bldg State: The Bldg Frontage is 35ft The Bldg Depth is 3(35ft) = 105ft 60’ 105’ 35’

Example  Golden Rectangle Let’s Revisit the Derivation of the GOLDEN RATIO A rectangle of length p and width q, with p > q, is called a golden rectangle if you can divide the rectangle into a square with side of length q and a smaller rectangle that is geometrically-similar to the original one. The GOLDEN RATION then = p/q

Example  Golden Rectangle Familiarize: Make a Diagram

Example  Golden Rectangle Translate: Use Diagram

Example  Golden Rectangle Carry Out: LET Φ ≡ Golden Ratio = p/q

Example  Golden Rectangle Carry Out: Since both p & q are distances they are then both POSITIVE Thus Φ = p/q must be POSITIVE State: GOLDEN RATIO as defined by the Golden Rectangle

Example  Pythagorus The hypotenuse of a right triangle is 52 yards long. One leg is 28 yards longer than the other. Find the lengths of the legs Familarize. First make a drawing and label it. Let s = length, in yards, of one leg. Then s + 28 = the length, in yards, of the other leg. 52 s s + 28

Pythagorean Triangle s + 28 s 52 Translate. We use the Pythagorean theorem: s2 + (s + 28)2 = 522 Carry Out. Identify the Quadratic Formula values a, b, & c s2 + (s + 28)2 = 522 s2 + s2 + 56s + 784 = 2704 2s2 + 56s − 1920 = 0 s2 + 28s − 960 = 0

Pythagorean Triangle Carry Out: With s2 + 28s − 960 = 0 Find: a = 1, b = 28, c = −960 Evaluate the Quadratic Formula

Pythagorean Triangle Carry Out: Continue Quadratic Eval s + 28 s 52

Pythagorean Triangle Check. Length cannot be negative, so −48 does not check. State. One leg is 20 yards and the other leg is 48 yards. 52 yds s = 20 yds s + 28yd = 48 yds

Vertical Ballistics In Physics 4A, you will learn the general formula for describing the height of an object after it has been thrown upwards: Where g ≡ the acceleration due to gravity A CONSTANT = 32.2 ft/s2 = 9.81 m/s2 t ≡ the time in flight, in s v0 ≡ the initial velocity in ft/s or m/s h0 ≡ the initial height in ft or m

Example  X-Games Jump In an extreme games competition, a motorcyclist jumps with an initial velocity of 80 feet per second from a ramp height of 25 feet, landing on a ramp with a height of 15 feet. Find the time the motorcyclist is in the air.

Example  X-Games Jump h = 15, v0 = 80, and h0 = 25 Familiarize: We are given the initial velocity, initial height, and final height and we are to find the time the bike is in the air. Can use the Ballistics Formula Translate: Use the formula with h = 15, v0 = 80, and h0 = 25. Use h = 15, v0 = 80, and h0 = 25

Example  X-Games Jump Carry Out: Use the Quadratic Formula to find t a = −16.1, b = 80 and c = 10

Example  X-Games Jump Carry Out: Since Times can NOT be Negative t ≈ 5.09 seconds Check: Check by substuting 5.09 for t in the ballistics Eqn. The Details are left for later State: The MotorCycle Flight-Time is very nearly 5.09 seconds

Find KW’s average speed Example  Biking Speed Kang Woo (KW to his friends) traveled by Bicycle 48 miles at a certain speed. If he had gone 4 mph faster, the trip would have taken 1 hr less. Find KW’s average speed

Example  Driving Speed Familiarize: In this can tabulating the information can help. Let r represent the rate, in miles per hour, and t the time, in hours for Kang Woo’s trip Distance Speed Time 48 r t r + 4 t – 1 Uses the Rate/Spd Eqn → Rate = Qty/Time

Example  Driving Speed Translate: Form the Table Obtain two Equations in r & t and Carry out: A system of equations has been formed. We substitute for r from the first equation into the second and solve the resulting equation

Example  Driving Speed Carry out: Next Clear Fractions Multiplying by the LCD

Example  Driving Speed Carry out: The Last Eqn is Quadratic in t: Solve by Quadratic Forumula with: a = 1, b = −1, and c = −12

Example  Driving Speed Carry out: By Quadratic Formula Since TIMES can NOT be NEGATIVE, then t = 4 hours Return to one of the table Eqns to find r 12 mph.

Example  Driving Speed Check: To see if 12 mph checks, we increase the speed 4 mph to16 mph and see how long the trip would have taken at that speed: The Answer checks a 3hrs is indeed 1hr less than 4 hrs that the trip actually took State: KW rode his bike at an average speed of 12 mph

ReCall The WORK Principle Suppose that A requires a units of time to complete a task and B requires b units of time to complete the same task. Then A works at a rate of 1/a tasks per unit of time. B works at a rate of 1/b tasks per unit of time, Then A and B together work at a rate of [1/a + 1/b] per unit of time.

The WORK Principle If A and B, working together, require t units of time to complete a task, then their rate is 1/t and the following equations hold:

Example  Empty Tower Tank A water tower has two drainpipes attached to it. Working alone, the smaller pipe would take 20 minutes longer than the larger pipe to empty the tower. If both drainpipes work together, the tower can be drained in 40 minutes. How long would it take the small pipe, working alone, to drain the tower?

Example  Empty Tower Tank Familiarize: Creating a Table helps to clarify the given data Pipe Time to Complete the Job Alone Work Rate Time Working Portion of Job Completed Smaller t + 20 min. 40 Larger t min. And the Job-Portions must add up to ONE complete Job:

Example  Empty Tower Tank Carry Out: The Last Eqn is Quadratic

Example  Empty Tower Tank Use Quadratic Formula:

Example  Empty Tower Tank Omit the negative solution as times cannot be negative The amount of time required by the small pipe is represented by t + 20, it would take approximately 20 + 71.2 or 91.2 minutes Check: Use the Work Eqn from Table 

Example  Empty Tower Tank State: Working alone the SMALL pipe would empty the Water Tower in about 91.2 minutes

WhiteBoard Work Problems From §8.2 Exercise Set 74 The Arrhenius Rate Equation

MotorCylcle Fatality Statistics All Done for Today MotorCylcle Fatality Statistics

Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu –

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